【leetcode】Best Time to Buy and Sell Stock III

Best Time to Buy and Sell Stock III

Say you have an array for which the i^th element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most two transactions.

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

 

设dpBack[i]为从第一天到第i天的最大的收益

设dpAfter[j]为从第j天到最后一天的最大收益

 

如何求最大收益可以考虑采用Best Time to Buy and Sell Stock I的方法

 

注意是最多买两次，也可以只买一次

 

 1 class Solution {

 2 public:

 3     int maxProfit(vector<int> &prices) {

 4        

 5         int n=prices.size();

 6         if(n==0)return 0;

 7  

 8         vector<int> dpBack(n),dpAfter(n);

 9  

10         int maxProfit=0;

11         dpBack[0]=0;

12         int left=prices[0];

13        

14         for(int i=1;i<n;i++)

15         {

16             if(prices[i]>left)

17             {

18                if(maxProfit<prices[i]-left) maxProfit=prices[i]-left;

19             }

20             else

21             {

22                 left=prices[i];

23             }

24            

25             dpBack[i]=maxProfit;

26         }

27        

28         maxProfit=0;

29         dpAfter[n-1]=0;

30         int right=prices[n-1];

31        

32         for(int j=n-2;j>=0;j--)

33         {

34             if(prices[j]<right)

35             {

36                 if(maxProfit<right-prices[j]) maxProfit=right-prices[j];

37             }

38             else

39             {

40                 right=prices[j];

41             }

42  

43             dpAfter[j]=maxProfit;

44         }

45        

46         int result=0;

47         for(int i=0;i<n-1;i++)

48         {

49             if(dpBack[i]+dpAfter[i+1]>result) result=dpBack[i]+dpAfter[i+1];

50            

51             if(result<dpBack[i+1]) result=dpBack[i+1];

52         }

53        

54         return result;

55     }

56 };